Question

A solenoid that is 66.2 cm long has a cross-sectional area of 18.0 cm2. There are 1300 turns of wire carrying a current of 8.15 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).

Answer 1

(a) Energy Density = 160.94 J/m³

(b) Energy Stored = 0.192 J

Step-by-step explanation:

(a)

The energy density of the magnetic field inside the solenoid is given by the following formula:

`Energy\ Denisty = (B^2)/(2\mu_o)\\`

where,

B = magnetic field strength of solenoid =
`(\mu_oNI)/(l)`

Therefore,

`Energy\ Density = (\mu_oN^2I^2)/(2l^2)`

where,

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

N = No. of turns = 1300

I = current = 8.15 A

L = length = 66.2 cm = 0.662 m

Therefore,

`Energy\ Density = ((4\pi\ x\ 10^(-7)\ N/A^2)(1300)^2(8.15\ A)^2)/(2(0.662\ m)^2)`

Energy Density = 160.94 J/m³

(b)

Energy Stored = (Energy Density)(Volume)

Energy Stored = (Energy Density)(Area)(L)

Energy Stored = (160.94 J/m³)(0.0018 m²)(0.662 m)

Energy Stored = 0.192 J